Question #fe939

1 Answer
Feb 1, 2018

#(2sqrt(3)+sqrt(6))/2#

Explanation:

If this is not the right expression, then obviously this will be wrong.

#3/(2sqrt(3)-sqrt(6))#

To rationalise the denominator we multiply the fraction by the conjugate of the denominator.

The conjugate of #(a+b)# is #(a-b)#. This has the property of giving the difference of 2 squares, which are always rational.

The conjugate of #(2sqrt(3)-sqrt(6))#, is #(2sqrt(3)+sqrt(6))#. Notice it is just a difference in the sign.

#:.#

#((2sqrt(3)+sqrt(6))*3)/((2sqrt(3)+sqrt(6))(2sqrt(3)-sqrt(6))#

#((2sqrt(3)+sqrt(6))*3)/(12-6)#

#((2sqrt(3)+sqrt(6))*3)/6#

#(2sqrt(3)+sqrt(6))/2#